Is $(2,5)$ the only solution?

Question

Find all pairs $(m,n)\in{\mathbb{N^2}}$ such that

$$(m^2-1)^3-n^2=2$$

Is $(2,5)$ the only solution?

Answer 1

Fermat claimed to have proved there is only one positive integer solution $x^3-y^2=2$, $(x,y)=(3,5)$. It is unknown if he actually had a proof.

The most common proof uses that $\mathbb Z[\sqrt{-2}]$ is a unique factorization domain.

That pretty much solves your question, setting $x=m^2-1,y=n$.

There might be an easier way to prove your subset has only one solution.

You could start by writing it as $$m^6-n^2=3(m^4-m^2+1)$$ or $$(m^2+1)(m^3-n)(m^3+n)=3(m^6+1)$$

Not sure where to go from there.

Answer 2

Maybe this helps

$(m^2-1)^3-n^2 =2\iff (m^2-1)^3-27=n^2-25$

$\iff (m^2-4)( (m^4-2m^2+1)+3(m^2-1)+9)=(n-5)(n+5)$

$\iff(m+2)(m-2)(m^4+m^2+7)=(n-5)(n+5)$